Given a random variable $T>0$ (time to event), I am using the following (general?) definition of the hazard function: $$ \lambda(t)=\lim_{h\to 0}\frac{\mathbb{P}(t\leq T < t + h \ | \ T \geq t)}{h} $$ according to this textbook. In this same text they split the continuous and discrete cases. For the continuous case they claim $$ \lambda(t) = \frac{f(t)}{S(t)} $$ where $f$ is the density and $S=1-F$ is the survival function. I derived this by \begin{aligned} \lambda(t) &= \lim_{h\to 0}\frac{\mathbb{P}(t\leq T < t + h\ | \ T \geq t)}{h} \\\\ &= \frac{1}{\mathbb{P}(T\geq t)} \cdot \lim_{h\to 0}\frac{\mathbb{P}(T<t+h)-\mathbb{P}(T<t)}{h} \\\\ &= \frac{1}{S(t)}\cdot\lim_{h\to 0}\frac{F(t+h)-F(t)}{h} \\\\ &= \frac{f(t)}{S(t)} \end{aligned} where we use the fact that strict and non-strict inequalities provide equal probabilities for continuous variables, and the definition of the derivative of the CDF.
However, I cannot seem to derive the simpler form for the discrete case ($T\in \{t_0,t_1,\dots \}$) in a similar fashion. It should become $$ \lambda(t_j) = \mathbb{P}(T=t_j \ | \ T\geq t_j) = \frac{p_j}{S(t_{j-1})} $$ where $p_j=\mathbb{P}(T=t_j)$. It makes a lot of sense and looks like a complete parallel to the continuous case, where now the division by the survival function is slightly different (1 time point behind) as in the discrete case, strict and non-strict inequalities do change matters slightly. When I try the following, it doesn't seem to work. \begin{align} \lim_{h\to 0}\frac{\mathbb{P}(t_j\leq T < t_j+h \ | \ T\geq t_j)}{h} &= \lim_{h\to 0}\frac{\mathbb{P}(t_j\leq T < t_j+h)}{\mathbb{P}(T\geq t_j)\cdot h} \\\\ &= \frac{1}{S(t_{j-1})}\cdot \lim_{h\to 0} \frac{\mathbb{P}(T < t_j+h)-\mathbb{P}(T < T_j)}{h} \end{align} From this it isn't so clear to me how to go to the PMF, as it was to get to the CDF/density in the continuous case. If I write (for small $h$, such that $t_j+h<t_{j+1}$) $$ \mathbb{P}(t_j\leq T < t_j+h) = \mathbb{P}(T = t_j) = p_j $$ it becomes $$ \lambda(t_j) = \lim_{h\to 0}\frac{p_j}{S(t_{j-1})\cdot h} $$ which the $h$ in the denominator blows up into $+\infty$. I think in this latter assumption for small $h$, it should perhaps somehow cancel the $h$ in the denominator so that the limit does not explode to $+\infty$.
Any ideas on how to get to the right conclusion, with the right steps? Am I missing something, or is the first 'general' definition of the hazard function only used for continuous variables, and the second outcome for the discrete case simply a separate definition of the hazard rate, which requires no proof?
